\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [378]

Optimal result

Integrand size = 26, antiderivative size = 146 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {32 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}+\frac {64 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}-\frac {48 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}+\frac {16 i (a+i a \tan (c+d x))^{9/2}}{9 a^8 d}-\frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^9 d} \] Output:

-32/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^5/d+64/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^6 
/d-48/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^7/d+16/9*I*(a+I*a*tan(d*x+c))^(9/2)/a 
^8/d-2/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^9/d
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \left (5419-6396 i \tan (c+d x)-4530 \tan ^2(c+d x)+1820 i \tan ^3(c+d x)+315 \tan ^4(c+d x)\right )}{3465 a^4 d} \] Input:

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

(2*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]*(5419 - (6396*I)*Tan[c + 
 d*x] - 4530*Tan[c + d*x]^2 + (1820*I)*Tan[c + d*x]^3 + 315*Tan[c + d*x]^4 
))/(3465*a^4*d)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3968, 53, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{10}}{(a+i a \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^4 \sqrt {i \tan (c+d x) a+a}d(i a \tan (c+d x))}{a^9 d}\)

\(\Big \downarrow \) 53

\(\displaystyle -\frac {i \int \left ((i \tan (c+d x) a+a)^{9/2}-8 a (i \tan (c+d x) a+a)^{7/2}+24 a^2 (i \tan (c+d x) a+a)^{5/2}-32 a^3 (i \tan (c+d x) a+a)^{3/2}+16 a^4 \sqrt {i \tan (c+d x) a+a}\right )d(i a \tan (c+d x))}{a^9 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \left (\frac {32}{3} a^4 (a+i a \tan (c+d x))^{3/2}-\frac {64}{5} a^3 (a+i a \tan (c+d x))^{5/2}+\frac {48}{7} a^2 (a+i a \tan (c+d x))^{7/2}+\frac {2}{11} (a+i a \tan (c+d x))^{11/2}-\frac {16}{9} a (a+i a \tan (c+d x))^{9/2}\right )}{a^9 d}\)

Input:

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((-I)*((32*a^4*(a + I*a*Tan[c + d*x])^(3/2))/3 - (64*a^3*(a + I*a*Tan[c + 
d*x])^(5/2))/5 + (48*a^2*(a + I*a*Tan[c + d*x])^(7/2))/7 - (16*a*(a + I*a* 
Tan[c + d*x])^(9/2))/9 + (2*(a + I*a*Tan[c + d*x])^(11/2))/11))/(a^9*d)
 

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(a^(m - 2)*b*f)   Subst[Int[(a - x)^(m/2 - 1)*(a + x 
)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && 
 EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
 

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {24 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{9}}\)	\(101\)
default	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {24 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{9}}\)	\(101\)

Input:

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^9*(-1/11*(a+I*a*tan(d*x+c))^(11/2)+8/9*a*(a+I*a*tan(d*x+c))^(9/2)- 
24/7*a^2*(a+I*a*tan(d*x+c))^(7/2)+32/5*a^3*(a+I*a*tan(d*x+c))^(5/2)-16/3*a 
^4*(a+I*a*tan(d*x+c))^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {64 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (128 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 704 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 1584 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 1848 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 1155 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{3465 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-64/3465*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(128*I*e^(11*I*d*x + 11 
*I*c) + 704*I*e^(9*I*d*x + 9*I*c) + 1584*I*e^(7*I*d*x + 7*I*c) + 1848*I*e^ 
(5*I*d*x + 5*I*c) + 1155*I*e^(3*I*d*x + 3*I*c))/(a^4*d*e^(10*I*d*x + 10*I* 
c) + 5*a^4*d*e^(8*I*d*x + 8*I*c) + 10*a^4*d*e^(6*I*d*x + 6*I*c) + 10*a^4*d 
*e^(4*I*d*x + 4*I*c) + 5*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {2 i \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 3080 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 11880 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 22176 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 18480 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4}\right )}}{3465 \, a^{9} d} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

-2/3465*I*(315*(I*a*tan(d*x + c) + a)^(11/2) - 3080*(I*a*tan(d*x + c) + a) 
^(9/2)*a + 11880*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 22176*(I*a*tan(d*x + c 
) + a)^(5/2)*a^3 + 18480*(I*a*tan(d*x + c) + a)^(3/2)*a^4)/(a^9*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,8192{}\mathrm {i}}{3465\,a^4\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,4096{}\mathrm {i}}{3465\,a^4\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{1155\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{693\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{99\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \] Input:

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^(7/2)),x)
 

Output:

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)* 
64i)/(11*a^4*d*(exp(c*2i + d*x*2i) + 1)^5) - ((a - (a*(exp(c*2i + d*x*2i)* 
1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*4096i)/(3465*a^4*d*(exp(c*2i 
+ d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d* 
x*2i) + 1))^(1/2)*1024i)/(1155*a^4*d*(exp(c*2i + d*x*2i) + 1)^2) - ((a - ( 
a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/( 
693*a^4*d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 
1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(99*a^4*d*(exp(c*2i + d*x*2i) 
 + 1)^4) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 
 1))^(1/2)*8192i)/(3465*a^4*d)
 

Reduce [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {too large to display} \] Input:

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

(sqrt(a)*( - 24*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**10*i - 17*int(( - s 
qrt(tan(c + d*x)*i + 1)*sec(c + d*x)**10*tan(c + d*x)**4)/(tan(c + d*x)**5 
*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c 
 + d*x)*i - 1),x)*tan(c + d*x)**2*d - 17*int(( - sqrt(tan(c + d*x)*i + 1)* 
sec(c + d*x)**10*tan(c + d*x)**4)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 
 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d + 48 
*int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**10*tan(c + d*x)**2)/(tan(c 
 + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 
 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d + 48*int(( - sqrt(tan(c + d* 
x)*i + 1)*sec(c + d*x)**10*tan(c + d*x)**2)/(tan(c + d*x)**5*i + 3*tan(c + 
 d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1) 
,x)*d - 14*int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**10*tan(c + d*x)) 
/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + 
d*x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d*i - 14*int(( - sqrt(t 
an(c + d*x)*i + 1)*sec(c + d*x)**10*tan(c + d*x))/(tan(c + d*x)**5*i + 3*t 
an(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)* 
i - 1),x)*d*i + int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**10)/(tan(c 
+ d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 
- 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d + int(( - sqrt(tan(c + d*x)*i 
 + 1)*sec(c + d*x)**10)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(...